well, i am not sure, but i think that factoring would really simplify things, because noteice that sin(npi), and cos(npi) are sonstants so you would end up with sth similar to
[tex]\int_0^{\pi} \frac{sin(n \cdot \pi) \cdot cos(y) + cos(n \pi) \cdot sin(y)}{y+n \pi}} dy=sin(n\cdot\pi)\int_0^{\pi}\frac{cos(y)}{y+n\pi}dy+cos(n\pi)\int_0^{\pi}\frac{sin(y)}{y+n\pi}dy= (-1)^{n} a_n[/tex]

well, i am not sure, but i think that factoring would really simplify things, because noteice that sin(npi), and cos(npi) are sonstants so you would end up with sth similar to
[tex]\int_0^{\pi} \frac{sin(n \cdot \pi) \cdot cos(y) + cos(n \pi) \cdot sin(y)}{y+n \pi}} dy=sin(n\cdot\pi)\int_0^{\pi}\frac{cos(y)}{y+n\pi}dy+cos(n\pi)\int_0^{\pi}\frac{sin(y)}{y+n\pi}dy[/tex]

now [tex]a_n = \int_0^{\pi} \frac{\sin y}{y + n\pi} \, dy[/tex]

Hi and thank you for your answer,

I can see that in order to achieve the result I need to get rid of this term here

[tex]sin(n\cdot\pi)\int_0^{\pi}\frac{cos(y)}{y+n\pi}dy[/tex]

Changing this into a sum would that allow me to get rid of this?

It can't simple be that I set n = 0 zero in the first part of the integral to get rid of it?